hdu 6038 Function【置换群进阶】

Function

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1389 Accepted Submission(s): 644

Problem Description
You are given a permutation a from 0 to n−1 and a permutation b from 0 to m−1.

Define that the domain of function f is the set of integers from 0 to n−1, and the range of it is the set of integers from 0 to m−1.

Please calculate the quantity of different functions f satisfying that f(i)=bf(ai) for each i from 0 to n−1.

Two functions are different if and only if there exists at least one integer from 0 to n−1 mapped into different integers in these two functions.

The answer may be too large, so please output it in modulo 109+7.

Input
The input contains multiple test cases.

For each case:

The first line contains two numbers n, m. (1≤n≤100000,1≤m≤100000)

The second line contains n numbers, ranged from 0 to n−1, the i-th number of which represents ai−1.

The third line contains m numbers, ranged from 0 to m−1, the i-th number of which represents bi−1.

It is guaranteed that ∑n≤106, ∑m≤106.

Output
For each test case, output “Case #x: y” in one line (without quotes), where x indicates the case number starting from 1 and y denotes the answer of corresponding case.

Sample Input
3 2
1 0 2
0 1
3 4
2 0 1
0 2 3 1

Sample Output
Case #1: 4
Case #2: 4

Source
2017 Multi-University Training Contest - Team 1

题意：
给定两个数列 a 和 b， 求有多少个 f 映射满足： f(i)=bf(ai)；
思路：
写了好久，结果样例画错一个群QAQ。要满足这种映射，来回之间的转换，需要用到置换群的思想，那么就需要求群和循环节，但是本题题意显然并不需要每个元素的循环节，而是把同在一个群的元素整合成一个群，这个群的长度便是这个群的循环节。合并之后，就要A映射B，当B序列某个群的循环节大于A序列某个群的循环节时，映射后的f函数最后一项是错的。换句话说，只有当B序列某个置换群循环节是A的因子，才能满足条件，B的这个子群的循环节就是此时满足映射关系的个数；

#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#define max_n 100010const int mod=1e9+7;using namespace std;typedef long long LL;int a[max_n],b[max_n],A[max_n],B[max_n],vis[max_n];int getp(int a[],int A[],int n) //求置换群的个数以及每个群的循环节 {    memset(vis,0,sizeof(vis));    int cnt=0;    for(int i=0;i<n;i++)    {        int t=a[i],res=0;//编号是0~n-1，所以起始res=0         if(!vis[i])        {            while(!vis[t])            {                vis[t]=1;                t=a[t];                res++;            }            A[cnt]=res; //每个群元素的个数也就是这个群的循环节             cnt++;  //群的个数         }    }    return cnt;}int main(){    int n,m;    while(scanf("%d %d",&n,&m)!=EOF)    {        memset(A,0,sizeof(A));        memset(B,0,sizeof(B));        for(int i=0;i<n;i++)            scanf("%d",&a[i]);        for(int i=0;i<m;i++)            scanf("%d",&b[i]);        int cnta=getp(a,A,n);        int cntb=getp(b,B,m);        LL ans=1;        for(int i=0;i<cnta;i++)        {            LL res=0;            for(int j=0;j<cntb;j++)            {                if(A[i]%B[j]==0)//当B序列是A序列的因子时，B序列长度就是方案数                     res+=B[j];            }            ans=(ans*res)%mod;        }        static int p=1;        printf("Case #%d: %lld\n",p++,ans);    }     return 0;}